If the plane P touches the sphere x² + y² + z² = r², then what is r equal to?

Correct Answer is : 529{5}{{29}}295

Correct Answer is : 529{5}{{29}}295

Test Index

UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 65-67

Direction: For the next three (3) items that follow:

A plane P passes through the line of intersection of the planes 2x – y + 3z = 2, x + y – z = 1 and the point (1, 0, 1).

Question : 67 of 120

Marks: +1, -0

x^{2} + y^{2} + z^{2} = r^{2}

r

equal to?

Solution:

Concept:

For a plane to touch a sphere, the perpendicular distance from the centre of the sphere to the plane must equal the radius of the sphere.

Explanation:

The sphere is given by

x^{2}+y^{2}+z^{2}=r^{2}

, so its centre is at

(0,0,0)

The plane is

5x+2y-5=0

Distance from centre

(0,0,0)

to the plane is:

d = \frac{\left|5(0)+2(0)-5\right|}{\sqrt{5^{2}+2^{2}+0^{2}}} = \frac{5}{\sqrt{29}}

Since the plane touches the sphere,

r = d

Therefore,

r = \frac{5}{\sqrt{29}}

Answer:

\frac{5}{\sqrt{29}}

(Option C)

Go to Question:

Prev Question

Next Question